Question: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $94$ years; the standard deviation is $18.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $75.8$ years.
Solution: $94$ $75.8$ $112.2$ $57.6$ $130.4$ $39.4$ $148.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $94$ years. We know the standard deviation is $18.2$ years, so one standard deviation below the mean is $75.8$ years and one standard deviation above the mean is $112.2$ years. Two standard deviations below the mean is $57.6$ years and two standard deviations above the mean is $130.4$ years. Three standard deviations below the mean is $39.4$ years and three standard deviations above the mean is $148.6$ years. We are interested in the probability of a turtle living longer than $75.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the turtles will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $75.8$ years and the other half $({16\%})$ will live longer than $112.2$ years. The probability of a particular turtle living longer than $75.8$ years is ${68\%} + {16\%}$, or $84\%$.